Locus of P such that the chord of contact of P with respect to `y^2= 4ax` touches the hyperbola `x^(2)-y^(2) = a^(2)` as

Options

A. `x^(2)+4y^(2)=4a^(2)`
B. `4x^(2)+y^(2)=4a^(2)`
C. `x^(2)+2y^(2)=2a^(2)`
D. `2x^(2)+y^(2)=2a^(2)`

Answer

To find the locus of the point \( P(x_1, y_1) \) such that the chord of contact with respect to the parabola \( y^2 = 4ax \) touches the hyperbola \( x^2 - y^2 = a^2 \), we can follow these steps:

### Step 1: Write the equation of the chord of contact
The chord of contact of the point \( P(x_1, y_1) \) with respect to the parabola \( y^2 = 4ax \) is given by:
\[ yy_1 = 2a(x + x_1) \]
This can be rearranged to:
\[ yy_1 = 2ax + 2ax_1 \]
or
\[ y = \frac{2a}{y_1}x + \frac{2ax_1}{y_1} \]

### Step 2: Write the equation of the tangent to the hyperbola
The equation of the tangent to the hyperbola \( x^2 - y^2 = a^2 \) in slope-intercept form is:
\[ y = mx + \sqrt{a^2m^2 - a^2} \]
where \( m \) is the slope of the tangent.

### Step 3: Set the slopes equal
Since the chord of contact touches the hyperbola, the slope of the chord of contact must equal the slope of the tangent to the hyperbola. From the chord of contact, the slope \( m \) is given by:
\[ m = \frac{2a}{y_1} \]
Thus, we have:
\[ \frac{2a}{y_1} = m \]

### Step 4: Substitute \( m \) into the tangent equation
Substituting \( m = \frac{2a}{y_1} \) into the tangent equation gives:
\[ y = \frac{2a}{y_1}x + \sqrt{a^2\left(\frac{2a}{y_1}\right)^2 - a^2} \]
This simplifies to:
\[ y = \frac{2a}{y_1}x + \sqrt{\frac{4a^4}{y_1^2} - a^2} \]

### Step 5: Equate the two expressions for \( y \)
Now, we equate the two expressions for \( y \):
\[ \frac{2a}{y_1}x + \frac{2ax_1}{y_1} = \frac{2a}{y_1}x + \sqrt{\frac{4a^4}{y_1^2} - a^2} \]
This implies:
\[ \frac{2ax_1}{y_1} = \sqrt{\frac{4a^4}{y_1^2} - a^2} \]

### Step 6: Square both sides
Squaring both sides results in:
\[ \left(\frac{2ax_1}{y_1}\right)^2 = \frac{4a^4}{y_1^2} - a^2 \]
This simplifies to:
\[ \frac{4a^2x_1^2}{y_1^2} = \frac{4a^4 - a^2y_1^2}{y_1^2} \]
Multiplying through by \( y_1^2 \) gives:
\[ 4a^2x_1^2 = 4a^4 - a^2y_1^2 \]

### Step 7: Rearranging the equation
Rearranging the equation yields:
\[ 4a^2x_1^2 + a^2y_1^2 = 4a^4 \]
Dividing through by \( a^2 \) gives:
\[ 4x_1^2 + y_1^2 = 4a^2 \]

### Step 8: Replace \( x_1 \) and \( y_1 \) with \( x \) and \( y \)
To find the locus, we replace \( x_1 \) and \( y_1 \) with \( x \) and \( y \):
\[ 4x^2 + y^2 = 4a^2 \]

### Final Answer
The locus of the point \( P \) is:
\[ \frac{x^2}{a^2} + \frac{y^2}{4a^2} = 1 \]
This represents an ellipse.